Let $(E, d)$ be a metric space and $\mathcal C$ the set of all non-empty compact subsets of $E$. We define a metric $d_H$ on $\mathcal C$ by $$ d_H (A, B) := \max \left \{ \max_{x \in B} d(x, A), \max_{y \in A} d (y, B) \right \}\quad \forall A,B \in \mathcal C, $$ with $d(x, A) := \inf_{y \in A} d(x, y)$. I'm trying to prove below property which I have seen from this Wikipedia page.
Theorem: If $E$ is complete, then so is $\mathcal C$.
Could you please have a check on my attempt?
My attempt: Assume $(A_n) \subset \mathcal C$ is a Cauchy sequence. Let $$ L := \{ \lim x_n \mid (x_n) \in (A_n)\text{ convergent}\}. $$
- $L$ is compact.
Let $(x_n) \subset L$. For each $n$, there is a sequence $(x_{n, m})_m \in (A_m)$ such that $x_{n, m} \xrightarrow{m \to \infty} x_n$. There is a subsequence $\varphi$ of $\mathbb N$ such that $d(x_{n, \varphi (n)}, x_n) < 1/n$. Let $y_n := x_{n, \varphi (n)} \in A_{\varphi (n)}$.
Lemma 1: Assume $(A_n) \subset \mathcal C$ is a Cauchy sequence and $(x_n) \in (A_n)$. Then $(x_n)$ has a Cauchy subsequence.
By our Lemma 1, there is a subsequence $\psi$ of $\mathbb N$ such that $(y_{\psi (n)})$ is Cauchy. Because $E$ is complete, $(y_{\psi (n)})$ is convergent. Hence $(x_{\psi (n)})$ is convergent.
- $A_n \to L$.
Assume the contrary that $A_n \not \to L$. Then there is $\varepsilon >0$ and a subsequence $\varphi$ of $\mathbb N$ such that $d_H (A_{\varphi (n)}, L) > \varepsilon$ for all $n$. There are $2$ cases.
- There exists a subsequence $\psi$ of $\varphi$ and a sequence $(a_{\psi (n)}) \in (A_{\psi (n)})$ such that $d (a_{\psi (n)}, L) > \varepsilon$ for all $n$.
By Lemma 1, there is a subsequence $\lambda$ of $\psi$ such that $(a_{\psi (n)}) \in (A_{\psi (n)})$ is Cauchy.
Lemma 2: Assume $(A_n) \subset \mathcal C$ is a Cauchy sequence, $\varphi$ a subsequence of $\mathbb N$, and $(x_{\varphi (n)}) \in (A_{\varphi (n)})$ a Cauchy sequence. Then there is a Cauchy sequence $(y_n) \in (A_n)$ such that $x_{\varphi (n)} = y_{\varphi (n)}$ for all $n$.
By Lemma 2, there is a Cauchy sequence $(y_n) \in (A_n)$ such that $a_{\psi (n)} = y_{\psi (n)}$ for all $n$. It follows that $\lim y_n =: \ell \in L$. Then there is $N$ such that $d(y_n, L) \le d(y_n, \ell) < \varepsilon$ for all $n >N$. There is $n_0$ such that $\varphi (n_0) > N$. Hence $d(a_{\varphi (n_0)}, L) = d(y_{\varphi (n_0)}, L) < \varepsilon$. This is a contradiction.
- There exists a subsequence $\psi$ of $\varphi$ and a sequence $(\ell_{\psi (n)}) \subset L$ such that $d (\ell_{\psi (n)}, A_{\psi (n)}) > \varepsilon$ for all $n$.
Because $L$ is compact, there is a subsequence $\lambda$ of $\psi$ and $\ell \in L$ such that $\ell_{\lambda (n)} \xrightarrow{n \to \infty} \ell$. Then $$ d(\ell, A_{\psi (n)}) \ge d (\ell_{\psi (n)}, A_{\psi (n)})- d(\ell, \ell_{\psi (n)}) > \varepsilon/2 $$ for $n$ large enough. Because $\ell \in L$, there is a sequence $(y_n) \in (A_n)$ such that $y_n \to \ell$. This implies $d(\ell, A_n) \to 0$, which is a contradiction. This completes the proof.
